You can see in each case that a 1.00 cm size cut-out gives the biggest volume of the open box. At first I used whole integers as my cut-out size, I then wanted to see if it was a smaller number that would give me my biggest volume, so I increased my sample size. Again it is easy to see that 1.00 cm still gives the biggest volume, but I wanted to go further, maybe it was a number like 0.95 or 1.05 and the computer was rounding up/down so I limited my sample size again. But still 1.00 cm cut-out gives us the biggest volume for our open topped box.
Algebra
Now we’ve got our first set of results and conclusive evidence to say that 1 cm cut-out to a 6cm x 6cm piece of card gives us the maximum volume for our open topped box we can use this to maybe find a formula. In this case we can see the equation of;
Volume = [(6 - 2 x cut-out) 2] x cut out
I could see a connection with 4 somewhere along the line, with both the width and depth being 4. You can also carry on the idea of relationship and connection through the 16 and 4, 16 being the squared from of 4. With the multiplying of the cut-out a relationship can be seen between the heights and cut-outs, there are the same. It is through the equation that everything is brought together and used to give the answer, just as it is in real life.
But you can also see where I get this from, from looking at the square we’re working on.
Here you can see the dimensions of a 6x6 cm cube. Knowing that to find the volume of a cube you multiply the height by the width by the depth. With C acting as the cut out with our square this would make;
6-2C x 6-2C x C
or
[(6-2C x C)2] x C
Limitations
- Results are for a 6x6cm card only.
- Only an approximation, to 2dp.
- We cannot predict the cut-out that gives max volume for all card sizes.
- We’ve only investigated a special type of rectangle- a square.
Plan
- Look at other pieces of square card.
- Look at answers to 3dp.
- Look at overall picture and try to find a rule that enables me to predict.
- Extend investigation.
Now I’m going to look at different square bits of card, and in accordance with the plan I’m now going to look as far as 3dp in my results.
5cm x 5cm piece of card.
With a 5cm x 5cm piece of card we’re able to see that a cut-out of 0.833 cm gives us the largest cut-out. A would not of necessarily been able to come to this conclusion if I don’t go ahead with a point that was brought forward in my plan and go to 3dp. If we now use the principle of my previous algebraic formula we can prove what we have seen.
V= [(5 – 2C) 2] x C
V= [(5 – 2x0.833) 2] x 0.833
V= 9.259258
It works, with graphing values into the formula we can see that we’ve got it right.
4cm x 4cm
With a 4cm x cm piece of card it is a cut-out of 0.667cm that gives us the maximum volume of 4.74074 cm3
V= [(4 – 2C) 2] x C
V= [(4 – 2x0.667) 2] x 0.677
V= 4.811815412
Although the calculation (and my calculator) give different answers, this could be because of simple problems and not because it’s wrong. In a situation like this, it usually comes down to decimal rounding and interpretation. Somewhere along the line either excel or my calculator have taken the decimal number and have changed this, because of this you get a differ of results, but close enough to see where your going. i.e turn the number into 1dp and you get 4.7 (1dp) which was the maximum original value to 1dp given when we looked at cut-out at 1dp.
3cm x 3cm
For a 3cm x 3cm piece of card we can clearly see that a cut out of 0.5cm gives us the maximum volume of a 2cm3.
V= [(3 - 2C) 2] x C
V= [(3 - 2x0.500) 2] x 0.500
V= 2
Again the formula seems to work and I have found the highest cut-out which will give the maximum volume.
Now we’ve we gone lower than a 6cm x 6cm piece of card, how about higher? Can the same process be used and the same formula called upon?
7cm x 7cm
You can see with these results, that the cut-out to a 7cm x 7cm piece of card that gives the maximum volume is 1.167cm, forming a volume of
25.40741 cm3 but do the previous ideas and formula work?
V= [(7 - 2C) 2] x C
V= [(7 - 2x1.167) 2] x 1.167
V= 25.407405852
This is what the results show and tell us, so the formula is correct.
Calculus
Calculus enables us to find the gradient function of a line. So instead of having to find the maximum volume with certain cut-out etc… we can draw the line of function and find the point of maximum volume together with its cut-out size. Differential calculus, an algebraic form, can help us to find the maximum cut-out quicker and easier.
e.g.
With differential calculus you can find out the gradient of where the maximum volume is produced from the cut-outs. The gradient at which the maximum cut-out can be found is when it is at 0, a straight line parallel to the x-axis.
Calculus for the 6cm x 6cm goes as follows;
We found that f(x) = (6-2x) (6-2x) x or as we know it V= [(6 – 2x)2]x
Which is f(x) = 36x-24x2+4x3
Turning that into f’(x) = 36-48x-12x2
This is also 12(3-4x+x2) = 0
Progression onto 12(x24x+3) = 0
Finishing with 12(x-3) (x-1) = 0
And for it to be true (i.e equal 0) c would have to equal 3 or 1. With a 6cm x 6cm this is in possible and therefore the maximum cut-out is 1, which is what we found out earlier. And as you can see from the graph above its right.
5cm x 5cm
We found that f(x) = (5-2x) (5-2x) x or as we know it V= [(5 – 2x)2]x
Which is f(x) = 25x-20x2+4x3
Turning that into f’(x) = 25-40x-12x2
This is also 12x2-40x+25 = 0
Progression onto 12x2-30x-10x+25 = 0
Finishing with (6x-5)(2x-5)
4cm x 4cm
We found that V= f(x) = (4-2x) (4-2x) x or as we know it V= [(4 – 2x)2]x
Which is f(x) = 16x-16x2-4x3
Turning that into f’(x) = 16-32x+12x2
This is also 4(4-8x+3x2) = 0
This can be then turned into 3x2-6x-2x+4 = 0
Making it into (3x-2)(x-2) = 0
Therefore with this the maximum cut-out for a 4cm x 4cm piece of card is
Cut-out = 2 or Cut-out = 2/3. With the answer found earlier being 0.667cm, 2/3 looks like the right answer.
But not only does this work with ones we already know, you can take it as far as to produce a general rule.
V=(a-2x)2x
V=(a2-4ax+4x2)x
Turning into ax2-4ax24x3
dv/dc = a2-8ax+12x2
When does the above equal 0?
Rearranged into; 12x2 -2ax-6ax+a2
This is then arranged into 2x(6x-a) – a(6x-a)
Gathered into (2x-a)(6x-a)=0
The cut-out is then either a/2 or a/6. With a divided by 2 Giving you the minimum, and a divided by 6 giving you the maximum.
Rectangles
I’ve done it with squares, and that seemed pretty simple, but now I’m going to look at rectangles. Will it be as simple? I’m going to look at rectangles with ratio 1:2. (Diagrams not drawn to scale)
To find out the volume of each one you can see that it’s going to follow the same kind of formula as the square apart from some alterations, with the rectangles it’s not the same 2 sides you’re multiplying, but 2 different ones. So for the 1cm x 2cm rectangle the equation for the volume would be as follows;
V= (1-2c)(2-2c)c
But this in itself may not help us, as from here you’ve got to go through every cut-out, then go to 2dp etc… Calculus can help us, finding the gradient function we can look at the graph, find where the gradient is 0 and that is our maximum cut-out, as well as what the volume itself is.
For the purpose of working we are to use a rectangle with sides A and KA (for you can always identify another side of a rectangle by it’s already known, or in this case, not known)
To find the volume of this rectangle when folded you would use this equation
V=(A-2C)(KA-2C)C
This can then be turned into
V=KA2C-2(A+KA)C2+4C3
This formula can now be used as a general formula for any piece of rectangular piece of card, it’s with this formula that we can plug in values and find maximum cut-out sizes and volumes to go with them. To get this kind of formula you can use calculus, to not only find a formula, but also to prove one.
If K=1 in the case of a square
V=A2C-2AC2+4C3
dv/dc = KA2-4(A+KA)C+12C2
When is dv/dc = 0? This meaning which cut-out ‘C’ gives max volume?
Gradient Function is 12C2-4A(1+K)C+KA2
Trying to factorise the equation in this state isn’t possible in the easiest and convenient way, therefore you have to use the equation;
The bits you plug into the formula you can find from this
ax2+bx+c
So with our formula
a=12 b=-4A(1+K) c=KA2 These are the co-efficient
You then get
Then
Then
Then
Then
Then
Then
This is now the function of a rectangle, in which you can find the cut-out of the rectangle which will give you the maximum volume, to prove this I’m going to look at when K=1.
And now I’m going to look at when K= 1/2 which is the way in which a rectangle is formed in the ration 2:1.
And it works, we find the cut-out to be of average 0.11A.
Conclusion
In this investigation I have been able to find out what and justify what I was originally set out to do, to find the maximum cut-out of any rectangular piece of card to give me the maximum volume. First of all I looked at square pieces of card and their actual measurements, volume after cat-out etc… From this information I was then able to look and concentrate on the measurements and come up with a general formula for a square piece of card, [(6-2C x C)2] x C I then went onto look at calculus with squares and was able to find out such things as the gradient function which is the moment where the gradient of the line, which the function belongs, is 0. Then I looked at rectangles with the ratio 1:2, forming a general formula again,
V= (1-2c)(2-2c)c for rectangles which I then used in calculus to help me find the answers to the volume with a k of 0.5. This algebraic way of working helped me to find a way in which I can find the cut-out of any piece of rectangular card which will give me the maximum volume.